The slope of the line graph of `log k` versus `1//T` for the reaction `N_(2)O_(5) rarr2NO_(2) + 1//2O_(2)` is `-5000`.Calculate the energy of activation of the reaction (in `kJ K^(-1) mol^(-1)`).

Options

A. `95.7`
B. `9.57`
C. `957`
D. None

Answer

A

Explanation

`K = Ae^(-E_(a)//RT)`
ln `K = ln A- (E_(a))/(RT)`
`2.303 logK = (E_(a))/(RT) + 2.303 logA`
`log K = (E_(a))/(2.303 RT) + logA`
Slope `(E_(a))/(2.303 R) = 5000`
`E_(a) = 5000 xx 8.31 xx 2.303`
`= 95.7 kJ k^(-1) mol^(-1)`